//
// Created by Jisam on 2024/8/11 23:14.
//
// `计算几何模板`
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//`Compares a double to zero`
int sgn(double x){
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}
/*
 * Point
 * Point()               - Empty constructor
 * Point(double _x,double _y)  - constructor
 * input()             - double input
 * output()            - %.2f output
 * operator ==         - compares x and y
 * operator <          - compares first by x, then by y
 * operator -          - return new Point after subtracting curresponging x and y
 * operator ^          - cross product of 2d points
 * operator *          - dot product
 * len()               - gives length from origin
 * len2()              - gives square of length from origin
 * distance(Point p)   - gives distance from p
 * operator + Point b  - returns new Point after adding curresponging x and y
 * operator * double k - returns new Point after multiplieing x and y by k
 * operator / double k - returns new Point after divideing x and y by k
 * rad(Point a,Point b)- returns the angle of Point a and Point b from this Point
 * trunc(double r)     - return Point that if truncated the distance from center to r
 * rotleft()           - returns 90 degree ccw rotated point
 * rotright()          - returns 90 degree cw rotated point
 * rotate(Point p,double angle) - returns Point after rotateing the Point centering at p by angle radian ccw
 */
struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){
        x = _x;
        y = _y;
    }
    void input(){
        scanf("%lf%lf",&x,&y);
    }
    void output(){
        printf("%.2f %.2f\n",x,y);
    }
    bool operator == (Point b)const{
        return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
    }
    bool operator < (Point b)const{
        return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;
    }
    Point operator -(const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    //叉积
    double operator ^(const Point &b)const{
        return x*b.y - y*b.x;
    }
    //点积
    double operator *(const Point &b)const{
        return x*b.x + y*b.y;
    }
    //返回长度
    double len(){
        return hypot(x,y);//库函数
    }
    //返回长度的平方
    double len2(){
        return x*x + y*y;
    }
    //返回两点的距离
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
    Point operator +(const Point &b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator *(const double &k)const{
        return Point(x*k,y*k);
    }
    Point operator /(const double &k)const{
        return Point(x/k,y/k);
    }
    //`计算pa  和  pb 的夹角`
    //`就是求这个点看a,b 所成的夹角`
    //`测试 LightOJ1203`
    double rad(Point a,Point b){
        Point p = *this;
        return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
    }
    //`化为长度为r的向量`
    Point trunc(double r){
        double l = len();
        if(!sgn(l))return *this;
        r /= l;
        return Point(x*r,y*r);
    }
    //`逆时针旋转90度`
    Point rotleft(){
        return Point(-y,x);
    }
    //`顺时针旋转90度`
    Point rotright(){
        return Point(y,-x);
    }
    //`绕着p点逆时针旋转angle`
    Point rotate(Point p,double angle){
        Point v = (*this) - p;
        double c = cos(angle), s = sin(angle);
        return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c);
    }
};
/*
 * Stores two points
 * Line()                         - Empty constructor
 * Line(Point _s,Point _e)        - Line through _s and _e
 * operator ==                    - checks if two points are same
 * Line(Point p,double angle)     - one end p , another end at angle degree
 * Line(double a,double b,double c) - Line of equation ax + by + c = 0
 * input()                        - inputs s and e
 * adjust()                       - orders in such a way that s < e
 * length()                       - distance of se
 * angle()                        - return 0 <= angle < pi
 * relation(Point p)              - 3 if point is on line
 *                                  1 if point on the left of line
 *                                  2 if point on the right of line
 * pointonseg(double p)           - return true if point on segment
 * parallel(Line v)               - return true if they are parallel
 * segcrossseg(Line v)            - returns 0 if does not intersect
 *                                  returns 1 if non-standard intersection
 *                                  returns 2 if intersects
 * linecrossseg(Line v)           - line and seg
 * linecrossline(Line v)          - 0 if parallel
 *                                  1 if coincides
 *                                  2 if intersects
 * crosspoint(Line v)             - returns intersection point
 * dispointtoline(Point p)        - distance from point p to the line
 * dispointtoseg(Point p)         - distance from p to the segment
 * dissegtoseg(Line v)            - distance of two segment
 * lineprog(Point p)              - returns projected point p on se line
 * symmetrypoint(Point p)         - returns reflection point of p over se
 *
 */
struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s = _s;
        e = _e;
    }
    bool operator ==(Line v){
        return (s == v.s)&&(e == v.e);
    }
    //`根据一个点和倾斜角angle确定直线,0<=angle<pi`
    Line(Point p,double angle){
        s = p;
        if(sgn(angle-pi/2) == 0){
            e = (s + Point(0,1));
        }
        else{
            e = (s + Point(1,tan(angle)));
        }
    }
    //ax+by+c=0
    Line(double a,double b,double c){
        if(sgn(a) == 0){
            s = Point(0,-c/b);
            e = Point(1,-c/b);
        }
        else if(sgn(b) == 0){
            s = Point(-c/a,0);
            e = Point(-c/a,1);
        }
        else{
            s = Point(0,-c/b);
            e = Point(1,(-c-a)/b);
        }
    }
    void input(){
        s.input();
        e.input();
    }
    void adjust(){
        if(e < s)swap(s,e);
    }
    //求线段长度
    double length(){
        return s.distance(e);
    }
    //`返回直线倾斜角 0<=angle<pi`
    double angle(){
        double k = atan2(e.y-s.y,e.x-s.x);
        if(sgn(k) < 0)k += pi;
        if(sgn(k-pi) == 0)k -= pi;
        return k;
    }
    //`点和直线关系`
    //`1  在左侧`
    //`2  在右侧`
    //`3  在直线上`
    int relation(Point p){
        int c = sgn((p-s)^(e-s));
        if(c < 0)return 1;
        else if(c > 0)return 2;
        else return 3;
    }
    // 点在线段上的判断
    bool pointonseg(Point p){
        return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
    }
    //`两向量平行(对应直线平行或重合)`
    bool parallel(Line v){
        return sgn((e-s)^(v.e-v.s)) == 0;
    }
    //`两线段相交判断`
    //`2 规范相交`
    //`1 非规范相交`
    //`0 不相交`
    int segcrossseg(Line v){
        int d1 = sgn((e-s)^(v.s-s));
        int d2 = sgn((e-s)^(v.e-s));
        int d3 = sgn((v.e-v.s)^(s-v.s));
        int d4 = sgn((v.e-v.s)^(e-v.s));
        if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
        return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
               (d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
               (d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
               (d4==0 && sgn((e-v.s)*(e-v.e))<=0);
    }
    //`直线和线段相交判断`
    //`-*this line   -v seg`
    //`2 规范相交`
    //`1 非规范相交`
    //`0 不相交`
    int linecrossseg(Line v){
        int d1 = sgn((e-s)^(v.s-s));
        int d2 = sgn((e-s)^(v.e-s));
        if((d1^d2)==-2) return 2;
        return (d1==0||d2==0);
    }
    //`两直线关系`
    //`0 平行`
    //`1 重合`
    //`2 相交`
    int linecrossline(Line v){
        if((*this).parallel(v))
            return v.relation(s)==3;
        return 2;
    }
    //`求两直线的交点`
    //`要保证两直线不平行或重合`
    Point crosspoint(Line v){
        double a1 = (v.e-v.s)^(s-v.s);
        double a2 = (v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
    }
    //点到直线的距离
    double dispointtoline(Point p){
        return fabs((p-s)^(e-s))/length();
    }
    //点到线段的距离
    double dispointtoseg(Point p){
        if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
            return min(p.distance(s),p.distance(e));
        return dispointtoline(p);
    }
    //`返回线段到线段的距离`
    //`前提是两线段不相交，相交距离就是0了`
    double dissegtoseg(Line v){
        return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
    }
    //`返回点p在直线上的投影`
    Point lineprog(Point p){
        return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
    }
    //`返回点p关于直线的对称点`
    Point symmetrypoint(Point p){
        Point q = lineprog(p);
        return Point(2*q.x-p.x,2*q.y-p.y);
    }
};
//圆
struct circle{
    Point p;//圆心
    double r;//半径
    circle(){}
    circle(Point _p,double _r){
        p = _p;
        r = _r;
    }
    circle(double x,double y,double _r){
        p = Point(x,y);
        r = _r;
    }
    //`三角形的外接圆`
    //`需要Point的+ /  rotate()  以及Line的crosspoint()`
    //`利用两条边的中垂线得到圆心`
    //`测试：UVA12304`
    circle(Point a,Point b,Point c){
        Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));
        Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));
        p = u.crosspoint(v);
        r = p.distance(a);
    }
    //`三角形的内切圆`
    //`参数bool t没有作用，只是为了和上面外接圆函数区别`
    //`测试：UVA12304`
    circle(Point a,Point b,Point c,bool t){
        Line u,v;
        double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.x);
        u.s = a;
        u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2));
        v.s = b;
        m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x);
        v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2));
        p = u.crosspoint(v);
        r = Line(a,b).dispointtoseg(p);
    }
    //输入
    void input(){
        p.input();
        scanf("%lf",&r);
    }
    //输出
    void output(){
        printf("%.2lf %.2lf %.2lf\n",p.x,p.y,r);
    }
    bool operator == (circle v){
        return (p==v.p) && sgn(r-v.r)==0;
    }
    bool operator < (circle v)const{
        return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0));
    }
    //面积
    double area(){
        return pi*r*r;
    }
    //周长
    double circumference(){
        return 2*pi*r;
    }
    //`点和圆的关系`
    //`0 圆外`
    //`1 圆上`
    //`2 圆内`
    int relation(Point b){
        double dst = b.distance(p);
        if(sgn(dst-r) < 0)return 2;
        else if(sgn(dst-r)==0)return 1;
        return 0;
    }
    //`线段和圆的关系`
    //`比较的是圆心到线段的距离和半径的关系`
    int relationseg(Line v){
        double dst = v.dispointtoseg(p);
        if(sgn(dst-r) < 0)return 2;
        else if(sgn(dst-r) == 0)return 1;
        return 0;
    }
    //`直线和圆的关系`
    //`比较的是圆心到直线的距离和半径的关系`
    int relationline(Line v){
        double dst = v.dispointtoline(p);
        if(sgn(dst-r) < 0)return 2;
        else if(sgn(dst-r) == 0)return 1;
        return 0;
    }
    //`两圆的关系`
    //`5 相离`
    //`4 外切`
    //`3 相交`
    //`2 内切`
    //`1 内含`
    //`需要Point的distance`
    //`测试：UVA12304`
    int relationcircle(circle v){
        double d = p.distance(v.p);
        if(sgn(d-r-v.r) > 0)return 5;
        if(sgn(d-r-v.r) == 0)return 4;
        double l = fabs(r-v.r);
        if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3;
        if(sgn(d-l)==0)return 2;
        if(sgn(d-l)<0)return 1;
    }
    //`求两个圆的交点，返回0表示没有交点，返回1是一个交点，2是两个交点`
    //`需要relationcircle`
    //`测试：UVA12304`
    int pointcrosscircle(circle v,Point &p1,Point &p2){
        int rel = relationcircle(v);
        if(rel == 1 || rel == 5)return 0;
        double d = p.distance(v.p);
        double l = (d*d+r*r-v.r*v.r)/(2*d);
        double h = sqrt(r*r-l*l);
        Point tmp = p + (v.p-p).trunc(l);
        p1 = tmp + ((v.p-p).rotleft().trunc(h));
        p2 = tmp + ((v.p-p).rotright().trunc(h));
        if(rel == 2 || rel == 4)
            return 1;
        return 2;
    }
    //`求直线和圆的交点，返回交点个数`
    int pointcrossline(Line v,Point &p1,Point &p2){
        if(!(*this).relationline(v))return 0;
        Point a = v.lineprog(p);
        double d = v.dispointtoline(p);
        d = sqrt(r*r-d*d);
        if(sgn(d) == 0){
            p1 = a;
            p2 = a;
            return 1;
        }
        p1 = a + (v.e-v.s).trunc(d);
        p2 = a - (v.e-v.s).trunc(d);
        return 2;
    }
    //`得到过a,b两点，半径为r1的两个圆`
    int gercircle(Point a,Point b,double r1,circle &c1,circle &c2){
        circle x(a,r1),y(b,r1);
        int t = x.pointcrosscircle(y,c1.p,c2.p);
        if(!t)return 0;
        c1.r = c2.r = r;
        return t;
    }
    //`得到与直线u相切，过点q,半径为r1的圆`
    //`测试：UVA12304`
    int getcircle(Line u,Point q,double r1,circle &c1,circle &c2){
        double dis = u.dispointtoline(q);
        if(sgn(dis-r1*2)>0)return 0;
        if(sgn(dis) == 0){
            c1.p = q + ((u.e-u.s).rotleft().trunc(r1));
            c2.p = q + ((u.e-u.s).rotright().trunc(r1));
            c1.r = c2.r = r1;
            return 2;
        }
        Line u1 = Line((u.s + (u.e-u.s).rotleft().trunc(r1)),(u.e + (u.e-u.s).rotleft().trunc(r1)));
        Line u2 = Line((u.s + (u.e-u.s).rotright().trunc(r1)),(u.e + (u.e-u.s).rotright().trunc(r1)));
        circle cc = circle(q,r1);
        Point p1,p2;
        if(!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2);
        c1 = circle(p1,r1);
        if(p1 == p2){
            c2 = c1;
            return 1;
        }
        c2 = circle(p2,r1);
        return 2;
    }
    //`同时与直线u,v相切，半径为r1的圆`
    //`测试：UVA12304`
    int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4){
        if(u.parallel(v))return 0;//两直线平行
        Line u1 = Line(u.s + (u.e-u.s).rotleft().trunc(r1),u.e + (u.e-u.s).rotleft().trunc(r1));
        Line u2 = Line(u.s + (u.e-u.s).rotright().trunc(r1),u.e + (u.e-u.s).rotright().trunc(r1));
        Line v1 = Line(v.s + (v.e-v.s).rotleft().trunc(r1),v.e + (v.e-v.s).rotleft().trunc(r1));
        Line v2 = Line(v.s + (v.e-v.s).rotright().trunc(r1),v.e + (v.e-v.s).rotright().trunc(r1));
        c1.r = c2.r = c3.r = c4.r = r1;
        c1.p = u1.crosspoint(v1);
        c2.p = u1.crosspoint(v2);
        c3.p = u2.crosspoint(v1);
        c4.p = u2.crosspoint(v2);
        return 4;
    }
    //`同时与不相交圆cx,cy相切，半径为r1的圆`
    //`测试：UVA12304`
    int getcircle(circle cx,circle cy,double r1,circle &c1,circle &c2){
        circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);
        int t = x.pointcrosscircle(y,c1.p,c2.p);
        if(!t)return 0;
        c1.r = c2.r = r1;
        return t;
    }

    //`过一点作圆的切线(先判断点和圆的关系)`
    //`测试：UVA12304`
    int tangentline(Point q,Line &u,Line &v){
        int x = relation(q);
        if(x == 2)return 0;
        if(x == 1){
            u = Line(q,q + (q-p).rotleft());
            v = u;
            return 1;
        }
        double d = p.distance(q);
        double l = r*r/d;
        double h = sqrt(r*r-l*l);
        u = Line(q,p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h)));
        v = Line(q,p + ((q-p).trunc(l) + (q-p).rotright().trunc(h)));
        return 2;
    }
    //`求两圆相交的面积`
    double areacircle(circle v){
        int rel = relationcircle(v);
        if(rel >= 4)return 0.0;
        if(rel <= 2)return min(area(),v.area());
        double d = p.distance(v.p);
        double hf = (r+v.r+d)/2.0;
        double ss = 2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
        double a1 = acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
        a1 = a1*r*r;
        double a2 = acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
        a2 = a2*v.r*v.r;
        return a1+a2-ss;
    }
    //`求圆和三角形pab的相交面积`
    //`测试：POJ3675 HDU3982 HDU2892`
    double areatriangle(Point a,Point b){
        if(sgn((p-a)^(p-b)) == 0)return 0.0;
        Point q[5];
        int len = 0;
        q[len++] = a;
        Line l(a,b);
        Point p1,p2;
        if(pointcrossline(l,q[1],q[2])==2){
            if(sgn((a-q[1])*(b-q[1]))<0)q[len++] = q[1];
            if(sgn((a-q[2])*(b-q[2]))<0)q[len++] = q[2];
        }
        q[len++] = b;
        if(len == 4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0)swap(q[1],q[2]);
        double res = 0;
        for(int i = 0;i < len-1;i++){
            if(relation(q[i])==0||relation(q[i+1])==0){
                double arg = p.rad(q[i],q[i+1]);
                res += r*r*arg/2.0;
            }
            else{
                res += fabs((q[i]-p)^(q[i+1]-p))/2.0;
            }
        }
        return res;
    }
};

/*
 * n,p  Line l for each side
 * input(int _n)                        - inputs _n size polygon
 * add(Point q)                         - adds a point at end of the list
 * getline()                            - populates line array
 * cmp                                  - comparision in convex_hull order
 * norm()                               - sorting in convex_hull order
 * getconvex(polygon &convex)           - returns convex hull in convex
 * Graham(polygon &convex)              - returns convex hull in convex
 * isconvex()                           - checks if convex
 * relationpoint(Point q)               - returns 3 if q is a vertex
 *                                                2 if on a side
 *                                                1 if inside
 *                                                0 if outside
 * convexcut(Line u,polygon &po)        - left side of u in po
 * gercircumference()                   - returns side length
 * getarea()                            - returns area
 * getdir()                             - returns 0 for cw, 1 for ccw
 * getbarycentre()                      - returns barycenter
 *
 */
struct polygon{
    int n;
    Point p[maxp];
    Line l[maxp];
    void input(int _n){
        n = _n;
        for(int i = 0;i < n;i++)
            p[i].input();
    }
    void add(Point q){
        p[n++] = q;
    }
    void getline(){
        for(int i = 0;i < n;i++){
            l[i] = Line(p[i],p[(i+1)%n]);
        }
    }
    struct cmp{
        Point p;
        cmp(const Point &p0){p = p0;}
        bool operator()(const Point &aa,const Point &bb){
            Point a = aa, b = bb;
            int d = sgn((a-p)^(b-p));
            if(d == 0){
                return sgn(a.distance(p)-b.distance(p)) < 0;
            }
            return d > 0;
        }
    };
    //`进行极角排序`
    //`首先需要找到最左下角的点`
    //`需要重载号好Point的 < 操作符(min函数要用) `
    void norm(){
        Point mi = p[0];
        for(int i = 1;i < n;i++)mi = min(mi,p[i]);
        sort(p,p+n,cmp(mi));
    }
    //`得到凸包`
    //`得到的凸包里面的点编号是0$\sim$n-1的`
    //`两种凸包的方法`
    //`注意如果有影响，要特判下所有点共点，或者共线的特殊情况`
    //`测试 LightOJ1203  LightOJ1239`
    void getconvex(polygon &convex){
        sort(p,p+n);
        convex.n = n;
        for(int i = 0;i < min(n,2);i++){
            convex.p[i] = p[i];
        }
        if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判
        if(n <= 2)return;
        int &top = convex.n;
        top = 1;
        for(int i = 2;i < n;i++){
            while(top && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0)
                top--;
            convex.p[++top] = p[i];
        }
        int temp = top;
        convex.p[++top] = p[n-2];
        for(int i = n-3;i >= 0;i--){
            while(top != temp && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0)
                top--;
            convex.p[++top] = p[i];
        }
        if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判
        convex.norm();//`原来得到的是顺时针的点，排序后逆时针`
    }
    //`得到凸包的另外一种方法`
    //`测试 LightOJ1203  LightOJ1239`
    void Graham(polygon &convex){
        norm();
        int &top = convex.n;
        top = 0;
        if(n == 1){
            top = 1;
            convex.p[0] = p[0];
            return;
        }
        if(n == 2){
            top = 2;
            convex.p[0] = p[0];
            convex.p[1] = p[1];
            if(convex.p[0] == convex.p[1])top--;
            return;
        }
        convex.p[0] = p[0];
        convex.p[1] = p[1];
        top = 2;
        for(int i = 2;i < n;i++){
            while( top > 1 && sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2])) <= 0 )
                top--;
            convex.p[top++] = p[i];
        }
        if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判
    }
    //`判断是不是凸的`
    bool isconvex(){
        bool s[2];
        memset(s,false,sizeof(s));
        for(int i = 0;i < n;i++){
            int j = (i+1)%n;
            int k = (j+1)%n;
            s[sgn((p[j]-p[i])^(p[k]-p[i]))+1] = true;
            if(s[0] && s[2])return false;
        }
        return true;
    }
    //`判断点和任意多边形的关系`
    //` 3 点上`
    //` 2 边上`
    //` 1 内部`
    //` 0 外部`
    int relationpoint(Point q){
        for(int i = 0;i < n;i++){
            if(p[i] == q)return 3;
        }
        getline();
        for(int i = 0;i < n;i++){
            if(l[i].pointonseg(q))return 2;
        }
        int cnt = 0;
        for(int i = 0;i < n;i++){
            int j = (i+1)%n;
            int k = sgn((q-p[j])^(p[i]-p[j]));
            int u = sgn(p[i].y-q.y);
            int v = sgn(p[j].y-q.y);
            if(k > 0 && u < 0 && v >= 0)cnt++;
            if(k < 0 && v < 0 && u >= 0)cnt--;
        }
        return cnt != 0;
    }
    //`直线u切割凸多边形左侧`
    //`注意直线方向`
    //`测试：HDU3982`
    void convexcut(Line u,polygon &po){
        int &top = po.n;//注意引用
        top = 0;
        for(int i = 0;i < n;i++){
            int d1 = sgn((u.e-u.s)^(p[i]-u.s));
            int d2 = sgn((u.e-u.s)^(p[(i+1)%n]-u.s));
            if(d1 >= 0)po.p[top++] = p[i];
            if(d1*d2 < 0)po.p[top++] = u.crosspoint(Line(p[i],p[(i+1)%n]));
        }
    }
    //`得到周长`
    //`测试 LightOJ1239`
    double getcircumference(){
        double sum = 0;
        for(int i = 0;i < n;i++){
            sum += p[i].distance(p[(i+1)%n]);
        }
        return sum;
    }
    //`得到面积`
    double getarea(){
        double sum = 0;
        for(int i = 0;i < n;i++){
            sum += (p[i]^p[(i+1)%n]);
        }
        return fabs(sum)/2;
    }
    //`得到方向`
    //` 1 表示逆时针，0表示顺时针`
    bool getdir(){
        double sum = 0;
        for(int i = 0;i < n;i++)
            sum += (p[i]^p[(i+1)%n]);
        if(sgn(sum) > 0)return 1;
        return 0;
    }
    //`得到重心`
    Point getbarycentre(){
        Point ret(0,0);
        double area = 0;
        for(int i = 1;i < n-1;i++){
            double tmp = (p[i]-p[0])^(p[i+1]-p[0]);
            if(sgn(tmp) == 0)continue;
            area += tmp;
            ret.x += (p[0].x+p[i].x+p[i+1].x)/3*tmp;
            ret.y += (p[0].y+p[i].y+p[i+1].y)/3*tmp;
        }
        if(sgn(area)) ret = ret/area;
        return ret;
    }
    //`多边形和圆交的面积`
    //`测试：POJ3675 HDU3982 HDU2892`
    double areacircle(circle c){
        double ans = 0;
        for(int i = 0;i < n;i++){
            int j = (i+1)%n;
            if(sgn( (p[j]-c.p)^(p[i]-c.p) ) >= 0)
                ans += c.areatriangle(p[i],p[j]);
            else ans -= c.areatriangle(p[i],p[j]);
        }
        return fabs(ans);
    }
    //`多边形和圆关系`
    //` 2 圆完全在多边形内`
    //` 1 圆在多边形里面，碰到了多边形边界`
    //` 0 其它`
    int relationcircle(circle c){
        getline();
        int x = 2;
        if(relationpoint(c.p) != 1)return 0;//圆心不在内部
        for(int i = 0;i < n;i++){
            if(c.relationseg(l[i])==2)return 0;
            if(c.relationseg(l[i])==1)x = 1;
        }
        return x;
    }
};
//`AB X AC`
double cross(Point A,Point B,Point C){
    return (B-A)^(C-A);
}
//`AB*AC`
double dot(Point A,Point B,Point C){
    return (B-A)*(C-A);
}
//`最小矩形面积覆盖`
//` A 必须是凸包(而且是逆时针顺序)`
//` 测试 UVA 10173`
double minRectangleCover(polygon A){
    //`要特判A.n < 3的情况`
    if(A.n < 3)return 0.0;
    A.p[A.n] = A.p[0];
    double ans = -1;
    int r = 1, p = 1, q;
    for(int i = 0;i < A.n;i++){
        //`卡出离边A.p[i] - A.p[i+1]最远的点`
        while( sgn( cross(A.p[i],A.p[i+1],A.p[r+1]) - cross(A.p[i],A.p[i+1],A.p[r]) ) >= 0 )
            r = (r+1)%A.n;
        //`卡出A.p[i] - A.p[i+1]方向上正向n最远的点`
        while(sgn( dot(A.p[i],A.p[i+1],A.p[p+1]) - dot(A.p[i],A.p[i+1],A.p[p]) ) >= 0 )
            p = (p+1)%A.n;
        if(i == 0)q = p;
        //`卡出A.p[i] - A.p[i+1]方向上负向最远的点`
        while(sgn(dot(A.p[i],A.p[i+1],A.p[q+1]) - dot(A.p[i],A.p[i+1],A.p[q])) <= 0)
            q = (q+1)%A.n;
        double d = (A.p[i] - A.p[i+1]).len2();
        double tmp = cross(A.p[i],A.p[i+1],A.p[r]) *
                     (dot(A.p[i],A.p[i+1],A.p[p]) - dot(A.p[i],A.p[i+1],A.p[q]))/d;
        if(ans < 0 || ans > tmp)ans = tmp;
    }
    return ans;
}

//`直线切凸多边形`
//`多边形是逆时针的，在q1q2的左侧`
//`测试:HDU3982`
vector<Point> convexCut(const vector<Point> &ps,Point q1,Point q2){
    vector<Point>qs;
    int n = ps.size();
    for(int i = 0;i < n;i++){
        Point p1 = ps[i], p2 = ps[(i+1)%n];
        int d1 = sgn((q2-q1)^(p1-q1)), d2 = sgn((q2-q1)^(p2-q1));
        if(d1 >= 0)
            qs.push_back(p1);
        if(d1 * d2 < 0)
            qs.push_back(Line(p1,p2).crosspoint(Line(q1,q2)));
    }
    return qs;
}
//`半平面交`
//`测试 POJ3335 POJ1474 POJ1279`
//***************************
struct halfplane:public Line{
    double angle;
    halfplane(){}
    //`表示向量s->e逆时针(左侧)的半平面`
    halfplane(Point _s,Point _e){
        s = _s;
        e = _e;
    }
    halfplane(Line v){
        s = v.s;
        e = v.e;
    }
    void calcangle(){
        angle = atan2(e.y-s.y,e.x-s.x);
    }
    bool operator <(const halfplane &b)const{
        return angle < b.angle;
    }
};
struct halfplanes{
    int n;
    halfplane hp[maxp];
    Point p[maxp];
    int que[maxp];
    int st,ed;
    void push(halfplane tmp){
        hp[n++] = tmp;
    }
    //去重
    void unique(){
        int m = 1;
        for(int i = 1;i < n;i++){
            if(sgn(hp[i].angle-hp[i-1].angle) != 0)
                hp[m++] = hp[i];
            else if(sgn( (hp[m-1].e-hp[m-1].s)^(hp[i].s-hp[m-1].s) ) > 0)
                hp[m-1] = hp[i];
        }
        n = m;
    }
    bool halfplaneinsert(){
        for(int i = 0;i < n;i++)hp[i].calcangle();
        sort(hp,hp+n);
        unique();
        que[st=0] = 0;
        que[ed=1] = 1;
        p[1] = hp[0].crosspoint(hp[1]);
        for(int i = 2;i < n;i++){
            while(st<ed && sgn((hp[i].e-hp[i].s)^(p[ed]-hp[i].s))<0)ed--;
            while(st<ed && sgn((hp[i].e-hp[i].s)^(p[st+1]-hp[i].s))<0)st++;
            que[++ed] = i;
            if(hp[i].parallel(hp[que[ed-1]]))return false;
            p[ed]=hp[i].crosspoint(hp[que[ed-1]]);
        }
        while(st<ed && sgn((hp[que[st]].e-hp[que[st]].s)^(p[ed]-hp[que[st]].s))<0)ed--;
        while(st<ed && sgn((hp[que[ed]].e-hp[que[ed]].s)^(p[st+1]-hp[que[ed]].s))<0)st++;
        if(st+1>=ed)return false;
        return true;
    }
    //`得到最后半平面交得到的凸多边形`
    //`需要先调用halfplaneinsert() 且返回true`
    void getconvex(polygon &con){
        p[st] = hp[que[st]].crosspoint(hp[que[ed]]);
        con.n = ed-st+1;
        for(int j = st,i = 0;j <= ed;i++,j++)
            con.p[i] = p[j];
    }
};
//***************************

const int maxn = 1010;
struct circles{
    circle c[maxn];
    double ans[maxn];//`ans[i]表示被覆盖了i次的面积`
    double pre[maxn];
    int n;
    circles(){}
    void add(circle cc){
        c[n++] = cc;
    }
    //`x包含在y中`
    bool inner(circle x,circle y){
        if(x.relationcircle(y) != 1)return 0;
        return sgn(x.r-y.r)<=0?1:0;
    }
    //圆的面积并去掉内含的圆
    void init_or(){
        bool mark[maxn] = {0};
        int i,j,k=0;
        for(i = 0;i < n;i++){
            for(j = 0;j < n;j++)
                if(i != j && !mark[j]){
                    if( (c[i]==c[j])||inner(c[i],c[j]) )break;
                }
            if(j < n)mark[i] = 1;
        }
        for(i = 0;i < n;i++)
            if(!mark[i])
                c[k++] = c[i];
        n = k;
    }
    //`圆的面积交去掉内含的圆`
    void init_add(){
        int i,j,k;
        bool mark[maxn] = {0};
        for(i = 0;i < n;i++){
            for(j = 0;j < n;j++)
                if(i != j && !mark[j]){
                    if( (c[i]==c[j])||inner(c[j],c[i]) )break;
                }
            if(j < n)mark[i] = 1;
        }
        for(i = 0;i < n;i++)
            if(!mark[i])
                c[k++] = c[i];
        n = k;
    }
    //`半径为r的圆，弧度为th对应的弓形的面积`
    double areaarc(double th,double r){
        return 0.5*r*r*(th-sin(th));
    }
    //`测试SPOJVCIRCLES SPOJCIRUT`
    //`SPOJVCIRCLES求n个圆并的面积，需要加上init\_or()去掉重复圆（否则WA）`
    //`SPOJCIRUT 是求被覆盖k次的面积，不能加init\_or()`
    //`对于求覆盖多少次面积的问题，不能解决相同圆，而且不能init\_or()`
    //`求多圆面积并，需要init\_or,其中一个目的就是去掉相同圆`
    void getarea(){
        memset(ans,0,sizeof(ans));
        vector<pair<double,int> >v;
        for(int i = 0;i < n;i++){
            v.clear();
            v.push_back(make_pair(-pi,1));
            v.push_back(make_pair(pi,-1));
            for(int j = 0;j < n;j++)
                if(i != j){
                    Point q = (c[j].p - c[i].p);
                    double ab = q.len(),ac = c[i].r, bc = c[j].r;
                    if(sgn(ab+ac-bc)<=0){
                        v.push_back(make_pair(-pi,1));
                        v.push_back(make_pair(pi,-1));
                        continue;
                    }
                    if(sgn(ab+bc-ac)<=0)continue;
                    if(sgn(ab-ac-bc)>0)continue;
                    double th = atan2(q.y,q.x), fai = acos((ac*ac+ab*ab-bc*bc)/(2.0*ac*ab));
                    double a0 = th-fai;
                    if(sgn(a0+pi)<0)a0+=2*pi;
                    double a1 = th+fai;
                    if(sgn(a1-pi)>0)a1-=2*pi;
                    if(sgn(a0-a1)>0){
                        v.push_back(make_pair(a0,1));
                        v.push_back(make_pair(pi,-1));
                        v.push_back(make_pair(-pi,1));
                        v.push_back(make_pair(a1,-1));
                    }
                    else{
                        v.push_back(make_pair(a0,1));
                        v.push_back(make_pair(a1,-1));
                    }
                }
            sort(v.begin(),v.end());
            int cur = 0;
            for(int j = 0;j < v.size();j++){
                if(cur && sgn(v[j].first-pre[cur])){
                    ans[cur] += areaarc(v[j].first-pre[cur],c[i].r);
                    ans[cur] += 0.5*(Point(c[i].p.x+c[i].r*cos(pre[cur]),c[i].p.y+c[i].r*sin(pre[cur]))^Point(c[i].p.x+c[i].r*cos(v[j].first),c[i].p.y+c[i].r*sin(v[j].first)));
                }
                cur += v[j].second;
                pre[cur] = v[j].first;
            }
        }
        for(int i = 1;i < n;i++)
            ans[i] -= ans[i+1];
    }
};